Finding Inverse Laplace Transform of a product series I need to compute the inverse Laplace transformation of the following equation. 
\begin{align*}
f(s)&=\frac{A}{\prod_{i=1}^{L}(s+a_i)^m} \\
&=\frac{A}{(s+a_1)^m\,(s+a_2)^m\cdots (s+a_L)^m}.
\end{align*}
What I managed to get is this:
$$\mathcal{L}^{-1}\!\left(\frac{1}{(s+a)^m}\right)=\frac{t^{m-1}e^{-at}}{{(m-1)!}}.$$
How do I find the Laplace inverse of the series though?
Someone suggested me using partial fraction decomposition and gave me this result: 
$$f(s)=\sum_{1\le k\le m\atop 1\le j\le L} {A_{jk}\over(s+a_j)^k},$$ 
but failed to tell how we achieved it.
 A: As you've mentioned, 
$$\mathcal{L}^{-1}\!\left(\frac{1}{(s+a)^m}\right)=\frac{t^{m-1}e^{-at}}{{(m-1)!}}$$
is the key to the whole business. If we can find the partial fraction decomposition of $f(s),$ we can invoke this result to get the final answer. Now, it is known in the theory of partial fraction decomposition that if you have a function like 
$$g(s)=\frac{1}{(s-1)^2(s+1)^2},$$
you can decompose it with the ansatz (an ansatz is an initial guess that we take as a working hypothesis):
$$g(s)=\frac{A}{s-1}+\frac{B}{(s-1)^2}+\frac{C}{s+1}+\frac{D}{(s+1)^2}. $$
If you do all the algebra, find common denominators, add everything up, and solve the resulting equations for $A,B,C,$ and $D,$ you find that
$$g(s)=-\frac{1}{4(s-1)}+\frac{1}{4(s-1)^2}+\frac{1}{4(s+1)}+\frac{1}{4(s+1)^2}.$$
So, we build up our solution as follows, bit-by-bit:
\begin{align*}
\frac{1}{(s+a_1)^m}&=\sum_{j=1}^{m}\frac{A_{1j}}{(s+a_1)^j} \\
\frac{1}{(s+a_2)^m}&=\sum_{j=1}^{m}\frac{A_{2j}}{(s+a_2)^j} \\
&\vdots \\
\frac{1}{(s+a_L)^m}&=\sum_{j=1}^{m}\frac{A_{Lj}}{(s+a_L)^j}.
\end{align*}
What we really want, of course, is all of these to be additive. We need to change the product in the original function entirely to a sum - which is what partial fraction decomposition is all about. So, we're going to write this (note the $A$ in the original can easily be absorbed into the $A_{ij}$):
\begin{align*}
f(s)&=\sum_{j=1}^{m}\frac{A_{1j}}{(s+a_1)^j}+\sum_{j=1}^{m}\frac{A_{2j}}{(s+a_2)^j}+\cdots + \sum_{j=1}^{m}\frac{A_{Lj}}{(s+a_L)^j} \\
&=\sum_{i=1}^L\sum_{j=1}^m\frac{A_{ij}}{(s+a_i)^j}.
\end{align*}
This is precisely what your friend you mentioned suggested to you. Now the final solution, once you find the $A_{ij},$ will then be
$$f(t)=\sum_{i=1}^L\sum_{j=1}^m\frac{A_{ij}\,t^{j-1}e^{-a_{i}t}}{(j-1)!}. $$
The major part of the work here, of course, is to find the $A_{ij}.$ I would recommend a computer algebra system for doing that, like Mathematica.
